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Quantum mechanics - Heisenberg picture

  1. Apr 18, 2017 #1
    1. The problem statement, all variables and given/known data
    We consider the O2- molecule, with the Hamiltonian and position operator having matrix representations in terms of the Pauli matrices:
    upload_2017-4-18_19-7-9.png

    In the Heisenberg picture, the position operator is:
    upload_2017-4-18_19-8-15.png

    (1) Find the eigenvalues and eigenstates of x(t) at time t=pi*hbar/(4A)

    (2) The state |-a> remains the same in the Heisenberg picture. Find the probabilities of measuring the two eigenvalues of x(t) at time t=pi*hbar/(4*A)

    (3) Now switching to the Schrodinger picture, find the state at time t = pi*hbar/(4A) for initial state |−a>

    (4) In the Schrodinger picture the operator x is constant. Find the probabilities of measuring eigenvalues +a and −a of x at time t = pi*hbar/(4*A), given initial state |−a>. Compare these to the result from question 2.

    2. Relevant equations
    The equations given above, plus the Born rule for calculating probabilities.

    3. The attempt at a solution
    (1) This part was ok, I got ± a for the eigenvalues, and the corresponding eigenvectors:
    (1/sqrt(2))*[1,∓ i]

    (2)
    the probabilities will be:
    [tex]
    \begin{equation}
    |<psi|\lambda_+>|^2
    \end{equation}
    [/tex]
    and
    [tex]
    \begin{equation}
    |<psi|\lambda_->|^2
    \end{equation}
    [/tex]
    where the lambdas are the eigenstates corresponding to the positive and negative eigenvalues. Psi is the state of the system, although I'm unsure of what it will be.

    (3)
    I'm not exactly sure how to interpret this question. Should I try to find an expression for psi in terms of |-a>?

    (4)
    Am I on the right track with the following?
    The probabilities will be:
    [tex]
    \begin{equation}
    |<-a|lambda_+>|^2
    \end{equation}
    [/tex]
    and
    [tex]
    \begin{equation}
    |<-a|lambda_->|^2
    \end{equation}
    [/tex]

    Any help is appreciated :)
     
  2. jcsd
  3. Apr 23, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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